--- title: "Transmission coefficient; 2nd order Butterworth LPF" subject: "Filter Design" chapter: 5 key: "semester_5/filter_design:transmission-coefficient-second-order" source: https://www.bctengineer.com/q/semester_5/filter_design/transmission-coefficient-second-order site: BCT Engineering --- # Transmission coefficient; 2nd order Butterworth LPF Filter Design · Chapter 5: Design of Resistively-Terminated Lossless Filters - Asked in: 2079 Baisakh, 1+5 marks; 2069 Chaitra, 1+1+5 marks (asked 2 times) - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/filter_design/transmission-coefficient-second-order - Whole chapter: https://www.bctengineer.com/md/semester_5/filter_design/chapter-5 ## Question What is transmission coefficient? What information do you get from the transmission coefficient? Design a second order Butterworth low pass filter using lossless ladder with equal termination of 1Ω i.e. R1 = 1Ω and R2 = 1Ω (Refer Table 1) ## Short answer Transmission coefficient |t|²=Pload/Pmax provides filter's power response, reaching1 at perfect match,0 at transmission zeros. Second-order Butterworth LPF, R1=R2=1 Ω: t=1/(s²+√2 s+1), ρ=s²/B(s), Z1 expansion yields L1=1.414 H, C2=1.414 F, R2=1 Ω (ωc=1 rad/s). ## Answer #### Transmission coefficient The **transmission coefficient** $t(s)$ is the ratio (in power terms) of the power actually delivered to the load $R_2$ to the maximum available power of the source $P_{\max} = |V_1|^2/4R_1$: $$ |t(j\omega)|^2 = \frac{P_2}{P_{\max}} = \frac{4R_1}{R_2}\left|\frac{V_2(j\omega)}{V_1(j\omega)}\right|^2 , \qquad t(s) = 2\sqrt{\frac{R_1}{R_2}}\,\frac{V_2(s)}{V_1(s)} $$ #### Information obtained from it - $|t(j\omega)|^2$ is the fraction of the available source power that reaches the load, so it directly gives the **filter's magnitude response** (passband, stopband, cutoff). - $|t| = 1$ means maximum power transfer (perfect match); $|t| = 0$ means a transmission zero (no power to the load). - Since $|t| \le 1$ for a passive lossless ladder, it fixes the maximum possible gain, e.g. $V_2/V_1 \le 1/2$ for equal terminations. - Through $|\rho|^2 = 1 - |t|^2$ it gives the reflection coefficient, which is used to find $Z_1$ and synthesize the ladder. #### Design of 2nd order Butterworth LPF ($R_1 = R_2 = 1\ \Omega$) **Step 1: Transmission coefficient.** For $n = 2$ (Table 1), $B(s) = s^2 + \sqrt2 s + 1$. With equal terminations $|t(0)|$ can be 1: $$ t(s) = \frac{1}{s^2+\sqrt2 s+1}, \qquad |t(j\omega)|^2 = \frac{1}{1+\omega^4} $$ (The actual voltage gain is $V_2/V_1 = \tfrac12 t(s)$, since $t = 2\sqrt{R_1/R_2}\,V_2/V_1$.) **Step 2: Reflection coefficient.** $$ |\rho(j\omega)|^2 = 1 - \frac{1}{1+\omega^4} = \frac{\omega^4}{1+\omega^4} \;\Rightarrow\; \rho(s)\rho(-s) = \frac{s^4}{B(s)B(-s)}, \quad \rho(s) = \frac{s^2}{s^2+\sqrt2 s+1} $$ **Step 3: Input impedance** ($R_1 = 1$): $$ Z_1(s) = \frac{1+\rho}{1-\rho} = \frac{2s^2+\sqrt2 s+1}{\sqrt2 s+1} $$ **Step 4: Continued fraction.** $$ \begin{aligned} Z_1 &= \sqrt2 s + \frac{1}{\sqrt2 s + 1} \\ &= 1.414\,s + \cfrac{1}{1.414\,s + 1} \end{aligned} $$ So series $L_1 = 1.414$ H, then a shunt $C_2 = 1.414$ F, and the last "1" is $R_2 = 1\ \Omega$. **Step 5: Circuit.** ```text R1=1 L1=1.414 H o-/\/\---UUU----+------+ | | | (~) V1 C2=1.414 F R2=1 | | | o---------------+------+ ``` **Answer:** $R_1 = 1\ \Omega$, $L_1 = 1.414$ H, $C_2 = 1.414$ F, $R_2 = 1\ \Omega$ (normalized, $\omega_c = 1$ rad/s). The dual form $Z_1 = (1-\rho)/(1+\rho)$ gives shunt $C_1 = 1.414$ F and series $L_2 = 1.414$ H. These agree with the standard table values (1.414, 1.414). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The transmission coefficient t(s) is related to the voltage ratio by: A. t = 2*sqrt(R1/R2)*V2/V1 B. t = V2/V1 C. t = sqrt(R1/R2)*V2/V1 D. t = R2/R1 * V2/V1 Answer: A. |t|^2 is the load power divided by the maximum available source power. (Key point: Definition of t) 2. |t| = 0 indicates: A. Maximum power transfer B. A transmission zero: no power reaches the load C. A perfect match D. A pole at the origin Answer: B. No power reaches the load at that frequency. (Key point: Information from t) 3. For n = 2, the reflection coefficient with R1 = R2 = 1 is: A. s^2/(s^2 + sqrt2 s + 1) B. 1/(s^2 + sqrt2 s + 1) C. s^2/(s^2 + 1) D. sqrt2 s/(s^2 + sqrt2 s + 1) Answer: A. |rho|^2 = w^4/(1+w^4). (Key point: Second-order Butterworth) 4. The ladder elements are: A. L1 = 1 H, C2 = 1 F B. L1 = 1.414 H, C2 = 1.414 F C. L1 = 2 H, C2 = 0.5 F D. L1 = 0.707 H, C2 = 0.707 F Answer: B. Z1 = 1.414 s + 1/(1.414 s + 1), matching standard table values. (Key point: Elements) ## Close variants set in other exams - [Transmission and reflection coefficients; 2nd order LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Atransmission-reflection-second-order) — Filter Design, chapter 5, asked 4× ## Related in Filter Design - [Zero reflection coefficient; 3rd order LPF equal termination](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Areflection-zero-third-order-equal) — Filter Design, chapter 5, asked 4× - [3rd order Butterworth LPF, R1 = 1 Ω, R2 = 4 Ω](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Abutterworth-lp-third-order-unequal-termination) — Filter Design, chapter 5, asked 4× - [Unity transmission coefficient; 3rd order LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Aunity-transmission-third-order-equal) — Filter Design, chapter 5, asked 3× - [Information from reflection coefficient; 3rd order LPF](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Areflection-info-third-order-equal) — Filter Design, chapter 5, asked 4× - [Reflection coefficient; 3rd order LPF with R2 = 4 Ω](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Areflection-coefficient-unequal-termination) — Filter Design, chapter 5, asked 7× - [Sallen-Key 2nd order Butterworth LPF design](https://www.bctengineer.com/question.md?key=semester_5%2Ffilter_design%3Asallen-key-butterworth-equal-resistors) — Filter Design, chapter 6, asked 3×